Determine if a Sudoku is valid, according to: .

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

 

A partially filled sudoku which is valid.

 

 

Note:

A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

 

tag :  HashSet 、  add 操作 及返回值。  

if(board[i][j] != '.' && !rows.add(board[i][j])) {

return false;

}

此行代码： 先做了添加操作，再返回的值，所以 二者合一。

另外：  而为矩阵的行列转换技巧  除法和求余数

 

public class Solution {    public boolean isValidSudoku(char[][] board) {                for(int i = 0; i < 9 ;i++) {            HashSet
     
       rows = new HashSet
      
       ();            HashSet
       
         cols = new HashSet
        
         ();            HashSet
         
           cube = new HashSet
          
           (); for(int j = 0; j < 9; j++) { //row if(board[i][j] != '.' && !rows.add(board[i][j])) { return false; } if(board[j][i] != '.' && !cols.add(board[j][i])) { return false; } int row = 3*(i/3); int col = 3*(i%3); if(board[row + j/3][col + j%3]!='.' && !cube.add(board[row + j/3][col + j%3])) return false; } } return true; }}